Question

T0 0 0 ] (1 point) The matrix A = -5 5 10 has two real eigenvalues, one of multiplicity 1 and one of multiplicity 2. Find the eigenvalues and a basis of [ 5 -5 -10] each eigenspace. 11 = has multiplicity 1, with a basis of 22 = !! has multiplicity 2, with a basis of 010 To enter a basis into WebWork, place the entries of each vector inside of brackets, and enter a list of these vectors, separated by commas. For instance, ( 11 11 if your basis is 31 2,1l, then you would enter [1,2,3],[1,1,1) into the answer blank. 311)

Answer 1

Solution: ro o 07 A = -5 5 10 and XE F = field 1 5 -5 -10 Now yet I be 3*3 Identity. Matrix 1A-111 - 1-1 0 0. 1-5 5-9 10 5 -5 -1007 26») (5-1) (-10-2) +504 ( Expanding along gst row) 3-1 F 12+54} = (-1) (a) (4+5) Bo 1A-111- 0 = 1=0,0,-5 Flow consider eigenspace; +1=0 ; Ca-oI) x = 0 - 1-5510 11% 15-5 40)/] co | 5*, -54%
1
So we have Бх, - Сх, + (0 x ) х = 2+23 — а , шү« Дели fot epub . Үte eigenvalue dao is НА "х, - І, І ду, - ~) (1) ( таһу х, Әо ) 4 0 1 - Take 8,20 | l in 0 / 402 Аав іумі 5 (A +51) х = 0 -) s o o fx, 1 -5 10 10 ч, 5 -5 -5 Лха . оо Съх, 1 - Бх, +1ox, 1 toҡа | (6xi -5 х = 5* ) х, о к.к. - - <3 а)
2
1. Thus 2 Required Eige vector Required en for A=-5 is fence - we have the eigenvalue =0 has multiplicity a, and basis for eigensface is and Eigenvalue y=-5 has multiplicity t, and basis for eigenspace is
3